\documentclass[paper=a4, fontsize=11pt]{scrartcl} % A4 paper and 11pt font size

\usepackage[T1]{fontenc} % Use 8-bit encoding that has 256 glyphs
\usepackage{fourier} % Use the Adobe Utopia font for the document - comment this line to return to the LaTeX default
\usepackage[english]{babel} % English language/hyphenation
\usepackage{amsmath,amsfonts,amsthm} % Math packages
\usepackage[UTF8]{ctex}
\usepackage{xcolor}
\usepackage{listings}
\usepackage{tikz}
\usepackage{lipsum} % Used for inserting dummy 'Lorem ipsum' text into the template
\usepackage{clrscode}
\usepackage{sectsty} % Allows customizing section commands
\usepackage[framed,numbered,autolinebreaks,useliterate]{mcode}
\usepackage{multirow}
\usepackage{graphicx}
\usepackage{float}
\usetikzlibrary{graphs}
\usetikzlibrary{shapes.arrows}
\allsectionsfont{\centering \normalfont\scshape} % Make all sections centered, the default font and small caps

\usepackage{fancyhdr} % Custom headers and footers
\pagestyle{fancyplain} % Makes all pages in the document conform to the custom headers and footers
\fancyhead{} % No page header - if you want one, create it in the same way as the footers below
\fancyfoot[L]{} % Empty left footer
\fancyfoot[C]{} % Empty center footer
\fancyfoot[R]{\thepage} % Page numbering for right footer
\renewcommand{\headrulewidth}{0pt} % Remove header underlines
\renewcommand{\footrulewidth}{0pt} % Remove footer underlines
\setlength{\headheight}{13.6pt} % Customize the height of the header

\numberwithin{equation}{section} % Number equations within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)
\numberwithin{figure}{section} % Number figures within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)
\numberwithin{table}{section} % Number tables within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)

\setlength\parindent{0pt} % Removes all indentation from paragraphs - comment this line for an assignment with lots of text

%----------------------------------------------------------------------------------------
%	TITLE SECTION
%----------------------------------------------------------------------------------------

\newcommand{\horrule}[1]{\rule{\linewidth}{#1}} % Create horizontal rule command with 1 argument of height

\title{
\normalfont \normalsize
\textsc{中国科学院大学}\ \textsc{计算机与控制学院} \\ [25pt] % Your university, school and/or department name(s)
\horrule{0.5pt} \\[0.4cm] % Thin top horizontal rule
\huge 矩阵应用与分析第四次作业 \\ % The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
}

\author{黎吉国&201618013229046} % Your name

\date{\normalsize\today} % Today's date or a custom date

\begin{document}

\maketitle % Print the title

\newpage
\section{answer for 4th}
$A(x,y,z)=(x+2y-z,-y,x+7z)$ is a linear operator on $\mathcal{R}^3$.
\begin{itemize}

\item (a) Determine $[A]_S$, where $S$ is the standard basis.
\item (b) Determine $[A]_{S'}$, as well as the nonsingular matrix $Q$ such that
\[ [A]_{S'}=Q^{-1}[A]_S Q \quad for \quad S'=\{ (1\ 0\ 0)^T,(1\ 1\ 0)^T,(1\ 1\ 1)^T \}. \]
\end{itemize}
\textbf{Solution:}\\
\begin{itemize}
\item (a)$S=\{e_1,e_2,e_3\}$\\
\begin{equation*}
  \begin{split}
A(e_1)&=A(1,0,0)=(1,0,1)=(1)e_1+(0)e_2+(1)e_3\\
A(e_2)&=A(0,1,0)=(2,-1,0)=(2)e_1+(-1)e_2+(0)e_3\\
A(e_3)&=A(0,0,1)=(-1,0,7)=(-1)e_1+(0)e_2+(7)e_3
\end{split}
\to
[A]_S=\left(
\begin{array}{rrr}
  1&2&-1\\
  0&-1&0\\
  1&0&7
\end{array}
\right)
\end{equation*}
\item (b) $S'=\{u_1,u_2,u_3\}=\{ (1\ 0\ 0)^T,(1\ 1\ 0)^T,(1\ 1\ 1)^T \}$
\begin{equation*}
\begin{split}
A(u_1)&=A(1,0,0)=(1,0,1)=(1)u_1+(-1)u_2+(1)u_3\\
A(u_2)&=A(1,1,0)=(3,-1,1)=(4)u_1+(-2)u_2+(1)u_3\\
A(u_3)&=A(1,1,1)=(2,-1,8)=(3)u_1+(-9)u_2+(8)u_3
\end{split}
\to
[A]_{S'}=\left(
\begin{array}{rrr}
  1&4&3\\
  -1&-2&-9\\
  1&1&8
\end{array}
\right)\\
\\
\begin{split}
Q&=([A(u_1)]_{S}|[A(u_2)]_{S}|[A(u_3)]_{S})\\
A(u_1)&=A(1,0,0)=(1,0,1)=(1)e_1+(0)e_2+(1)e_3\\
A(u_2)&=A(1,1,0)=(3,-1,1)=(3)e_1+(-1)e_2+(1)e_3\\
A(u_3)&=A(1,1,1)=(2,-1,8)=(2)e_1+(-1)e_2+(8)e_3
\end{split}
\to
Q&=\left(
\begin{array}{rrr}
  1&3&2\\
  0&-1&-1\\
  1&1&8
\end{array}
\right)\\
[A]_{S'}=Q^{-1}[A]_{S}Q
\end{equation*}
\end{itemize}

\newpage
\section{answer for 7th}
Let $T$ be  the linear operator on $\mathcal{R}^4$ defined by
\[ T(x_1,x_2,x_3,x_4)=(x_1+x_2+2x_3-x_4, x_2+x_4, 2x_3-x_4, x_3+x_4) \]
and let $\mathcal{X}=span\{e_1, e_2\}$ be the subspace that is spenned by the first two unit vector in $\mathcal{R}^4$.
\begin{itemize}
\item (a) Explain why $\mathcal{X}$ is invariant under $T$.
\item (b) Determine $[T_{/\mathcal{X}}]_{\{e_1,e_2\}}$.
\item (c) Describe the structure of $[T]_{\beta}$,where $\beta$ is any basis obtained from an extension of $\{e_1,e_2\}$.
\end{itemize}
\textbf{Solution:}
\begin{itemize}
\item (a) $T(e_1)=T(1,0,0,0)=(1,0,0,0)=(1)e_1$,$T(e_2)=T(0,1,0,0)=(1,1,0,0)=(1)e_1+(1)e_2$\\
for any vector $x\in \mathcal{X},x=\alpha e_1+\beta e_1$,
\[ T(x)=T(\alpha e_1+ \beta e_2)=\alpha T(e_1) + \beta T(e_2)=(\alpha+\beta)e_1+\beta e_2 \in \mathcal{X} \]
so $\mathcal{X}$ is a invariant subspace.
\item (b)
\begin{equation*}
[T_{/\mathcal{X}}]_{e_1,e_2}=( [T_{/\mathcal{X}}(e_1)]_{\beta} | [T_{/\mathcal{X}}(e_2)]_{\beta}  )=
\left(
\begin{array}{rr}
  1&0\\
  0&1
\end{array}
\right)
\end{equation*}
\item 
$$

\end{itemize}

\end{document}
